Difference between revisions of "2009 AIME I Problems/Problem 5"
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== Solution == | == Solution == | ||
+ | Sorry, I fail to get the diagram up here, someone help me. | ||
+ | |||
+ | Since <math>K</math> is the midpoint of <math>PM, AC</math>. | ||
+ | |||
+ | Thus, <math>AK=CK,PK=MK</math> and the opposite angles are congruent. | ||
+ | |||
+ | Therefore, triangle <math>AMK</math> is congruent to triangle <math>CPK</math> | ||
+ | |||
+ | angle <math>KMA</math> is congruent to <math>KPA</math> because of CPCTC | ||
+ | |||
+ | That shows <math>AM</math> is parallel to <math>CP</math> (also <math>CL</math>) | ||
+ | |||
+ | That makes triangle <math>AMB</math> congruent to <math>LPB</math> | ||
+ | |||
+ | Thus, <math>\frac {AM}{LP}=\frac {AB}{LB}</math> | ||
+ | |||
+ | <cmath>\frac {AM}{LP}=\frac {AB}{LB}=\frac {AL+LB}{LB}=\frac {AL}{LB}+1</cmath> | ||
+ | |||
+ | Now let apply angle bisector thm. | ||
+ | |||
+ | <cmath>\frac {AL}{LB}=\frac {AC}{BC}=\frac {450}{300}=\frac {3}{2}</cmath> | ||
+ | |||
+ | <cmath>\frac {AM}{LP}=frac {AL}{LB}+1=frac {5}{2}</cmath> | ||
+ | |||
+ | <cmath>\frac {180}{LP}=frac {5}{2}</cmath> | ||
+ | |||
+ | <cmath>LP=\boxed {072}</cmath> | ||
== See also == | == See also == | ||
{{AIME box|year=2009|n=I|num-b=4|num-a=6}} | {{AIME box|year=2009|n=I|num-b=4|num-a=6}} |
Revision as of 23:31, 20 March 2009
Problem
Triangle has
and
. Points
and
are located on
and
respectively so that
, and
is the angle bisector of angle
. Let
be the point of intersection of
and
, and let
be the point on line
for which
is the midpoint of
. If
, find
.
Solution
Sorry, I fail to get the diagram up here, someone help me.
Since is the midpoint of
.
Thus, and the opposite angles are congruent.
Therefore, triangle is congruent to triangle
angle is congruent to
because of CPCTC
That shows is parallel to
(also
)
That makes triangle congruent to
Thus,
Now let apply angle bisector thm.
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |